Projectile motioncan be described byvertical componentsandhorizontal componentsof motion.

Unit 2: Projectile Motion

Vectors and Scalars

Honors Physics

Scalar

A SCALAR is ANY quantity in physics that has MAGNITUDE, but has NO direction associated with it.

Magnitude – A numerical value with units.

Scalar Example

Magnitude

Speed 20 m/s

Distance 10 m

Age 15 years

Heat1000

calories

Vector

A VECTOR is ANY quantity in physics that has BOTH MAGNITUDE and DIRECTION.

VectorMagnitude & Direction

Velocity 20 m/s, N

Acceleration 10 m/s/s, E

Force 5 N, West

Faxv

,,,Vectors are typically illustrated by drawing an ARROW above the symbol. The arrow is used to convey direction and magnitude.

Applications of Vectors

VECTOR ADDITION – If 2 similar vectors point in the SAME direction, add them.

Example: A person travels 54.5 meters east, then another 30 meters easterly. Calculate the displacement relative to starting point.

54.5 m, E 30 m, E+

84.5 m, E

Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION.

Applications of Vectors

VECTOR SUBTRACTION - If 2 vectors are going in opposite directions, you SUBTRACT.

Example: A man walks 54.5 meters east, then 30 meters west. Calculate his displacement relative to where he started?

54.5 m, E

30 m, W-

24.5 m, E

Non-Collinear VectorsWhen 2 vectors are perpendicular, you must use

the Pythagorean theorem.

95 km,E

55 km, N

Start

Finish

A man walks 95 km, East then 55 km, north. Calculate his RESULTANT DISPLACEMENT.The hypotenuse in Physics

is called the RESULTANT.

The LEGS of the triangle are called the COMPONENTS

Horizontal Component

Vertical Component kmc

c

bacbac

8.10912050

5595Resultant 22

22222

BUT……what about the direction?In the previous example, DISPLACEMENT was asked for

and since it is a VECTOR we should include a DIRECTION on our final answer.

NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE.

N

S

EW

N of E

E of N

S of W

W of S

N of W

W of N

S of E

E of S

N of E

BUT...what about the ANGLE VALUE..?Just putting North of East on the answer is NOT specific enough for the direction. We MUST find the VALUE of the angle.

N of E

55 km, N

95 km,E

To find the value of the angle we use a Trig function called TANGENT.

30)5789.0(

5789.095

55

1

Tan

sideadjacent

sideoppositeTan

109.8 km

So the COMPLETE final answer is : 109.8 km, 30 degrees North of East

What if you are missing a component?Suppose a person walked 65 m, 25 degrees East of North. What

were his horizontal and vertical components?

65 m25

H.C. = ?

V.C = ?

The goal: ALWAYS MAKE A RIGHT TRIANGLE!

To solve for components, we often use the trig functions tan, sin and cosine.

EmCHopp

NmCVadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,47.2725sin65..

,91.5825cos65..

sincos

sincosine

ExampleA bear, searching for food wanders 35 meters east then 20 meters north. Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement.

35 m, E

20 m, N

12 m, W

6 m, S

- =23 m, E

- =14 m, N

23 m, E

14 m, N

3.31)6087.0(

6087.23

14

93.262314

1

22

Tan

Tan

mR

The Final Answer: 26.93 m, 31.3 degrees NORTH of EAST

R

ExampleA boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.

15 m/s, N

8.0 m/s, W

Rv

1.28)5333.0(

5333.015

8

/17158

1

22

Tan

Tan

smRv

The Final Answer : 17 m/s, @ 28.1 degrees West of North

ExampleA plane moves with a velocity of 63.5 m/s at 32o South of East. Calculate the plane's horizontal and vertical velocity components.

63.5 m/s

32

H.C. =?

V.C. = ?

SsmCVopp

EsmCHadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,/64.3332sin5.63..

,/85.5332cos5.63..

sincos

sinecosine

ExampleA storm system moves 5000 km due east, then shifts course at 40 degrees North of East for 1500 km. Calculate the storm's resultant displacement.

NkmCVopp

EkmCHadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,2.96440sin1500..

,1.114940cos1500..

sincos

sinecosine

5000 km, E

40

1500 km

H.C.

V.C.

5000 km + 1149.1 km = 6149.1 km

6149.1 km

964.2 kmR

91.8)1568.0(

1568.01.6149

2.964

1.62242.9646149

1

22

Tan

Tan

kmR

The Final Answer: 6224.1 km @ 8.91 degrees, N of E

MINUTE PHYSICS

We’ve seen simple straight-line motion

(linear )

Now, apply these ideas to curved motion

(nonlinear)

A combination of horizontal and vertical

motion.

Unit 2B: Projectile Motion

vector quantity:

velocity (v)

3.1 Vector and Scalar Quantities

Scalar quantity has magnitude only

Vector quantity has magnitude and direction

size, length, ...

scalar quantity:

speed

80 km/h north

80 km/h

acceleration (a)?

A plane’s velocity is often the result of combining two or more other velocities. • a small plane flies north at 80 km/h• a tailwind blows north at 20 km/h

3.2 Velocity Vectors

80 km/h

20 km/h

100 km/h

20 km/h

80 km/h

60 km/h

What if the plane flies against the wind?

vector addition:

same direction (ADD)

opp. direction (SUB)

Consider a plane flying 80 km/h north, but… caught in a strong crosswind of 60 km/h east.

The two velocity vectors must be combined to find the resultant.

3.2 Velocity Vectors

80 km/h

60 km/h

resultant

An 80 km/h plane flyingin a 60 km/h crosswindhas a resultant speed of 100 km/h relative to the ground.

HOW?

100 km/h

Consider a plane flying 80 km/h north, but… caught in a strong crosswind of 60 km/h east.

The two velocity vectors must be combined to find the resultant.

3.2 Velocity Vectors

80 km/h

60 km/h

1) draw vectors tail-to-head.

2) a2 + b2 = c2

vector addition:

(80)2 + (60)2 = c2

√(6400 + 3600) = c

100 km/h

The 80 km/h and 60 km/h vectors produce aresultant vector of 100 km/h, but…in what direction?

3.2 Velocity Vectors

80 km/h

60 km/h

100 km/h

θ = tan-1(opp/adj)

tan(θ) = opp/adj

θθ : “theta”

θ = 53o N of E

100 km/h, 53o N of E(or 53o above + x-axis)

opp

adjθ = tan-1(80/60)

Suppose that an airplane normally flying at80 km/h encounters wind at a right angle toits forward motion—a crosswind.

Will the airplane fly faster or slower than80 km/h?

Answer:A crosswind would increase the speed of the airplane but blow it off course by a predictable amount.

3.2 Velocity Vectors

1. Which of these expresses a vector quantity?

A. 10 kg

B. 10 kg to the north

C. 10 m/s

D. 10 m/s 23o N of E

Quick Quiz!

3.1

2. An ultra-light aircraft traveling north at 40 km/h in a 30 km/h crosswind (at right angles) has a groundspeed of _____.

A. 30 km/h

B. 40 km/h

C. 50 km/h

D. 60 km/h

Quick Quiz.

3.2

Check off the learning targets you can do after today.

40 km/h

30 km/h

??? km/h

a2 + b2 = c2

(30)2 + (40)2 = c2

√(900 + 1600) = c

You can resolve a single vector into two component vectors at right angles to each other:

3.3 Components of Vectors

Vectors X and Y are the horizontal and vertical components of a vector V.

A ball’s velocity can be resolved into horizontal (x) and vertical (y) components.

3.3 Components of Vectors

3.3 Components of Vectors

A jet flies 340 m/s (mach 1) at 60o N of E.What are the vertical and horizontal components of the jet’s velocity?

vy = ?

vx = ?

60o

vy

vx

opp

adj

340 m/s

vy = v sin(θ)

sin(θ) = opp/hypcos(θ) = adj/hyp

vx = v cos(θ)

(hyp)

(v)

3.3 Components of Vectors

A jet flies 340 m/s (mach 1) at 60o N of E.What are the vertical and horizontal components of the jet’s velocity?

vy = ?

vx = ?

60o

vy

vx

opp

adj

340 m/s

vy = v sin(θ)

vx = v cos(θ)

(hyp)

(v)

vy = (340 m/s) • sin(60)

vy =

vx = (340 m/s) • cos(60)

vx =

294 m/s

170 m/s

294 m/s

170 m/s

1. A ball launched into the air at 45° to the horizontal initially has…

A. equal horizontal and vertical components.

B. components that do not change in flight.

C.components that affect each other throughout flight.

D.a greater component of velocity than the vertical component.

Quick Quiz!

3.3

30o

2. A jet flies 680 m/s (mach 2) at 30o N of E. What is the vertical component of the jet’s velocity (vy)?

A. 589 m/s

B. 340 m/s

C.230 m/s

D.180 m/s

Quick Quiz.

680 m/s

vy = v sin(θ) = opp / hyp

vy = (680 m/s) • sin(30)

340 m/s

projectile:

any object moving through a path, acted on only by gravity. (no friction/no air resistance)Ex: cannonball, ball/stone, spacecraft/satellite, etc.

3.4 Projectile Motion

projectile motion

gravity-free path

gravity only

Projectile motion is separated into components.

a. Roll a ball horizontally, v is constant, b/c no acceleration from g horizontally.

b. Drop a ball, it accelerates downward covering a greater distance each second.

c. x & y components are completely independent of each other.

3.4 Projectile Motion

Projectile motion is separated into components.

a. Roll a ball horizontally, v is constant, b/c no acceleration from g horizontally.

b. Drop a ball, it accelerates downward covering a greater distance each second.

c. x & y components are completely independent of each other.

d. combined they cause curved paths.

3.4 Projectile Motion

• x component is constant (a = 0)(g acts only in y direction)

• both fall the same y distance in same time.(x and y are completely unrelated)

3.4 Projectile Motion

vx vy

3.4 Projectile Motion

vx2 vx4

vy2

vy4

vx3

vy3

1. When no air resistance acts on a fast-moving baseball, its acceleration is …

A. downward only

B. in the forward x direction it was thrown

C. opposite to the force of gravity

D. both forward and downward

Quick Quiz!

The Y distance fallen is the same vertical distance it would fall if dropped from rest.

Remember d = ½gt2

3.5 Projectiles Launched at an Angle

Height & Rangevx is constant, but vy changes.

At the max height, vy = 0.(only Vx)

3.5 Projectiles Launched at an Angle

launch angle affects height (y) and range (x)

3.5 Projectiles Launched at an Angle

Height & Range

height

range

height

range

60o 75o

more angle:

-more initial vy, more height

-less initial vx, less range

• angles that add to 90° have equal ranges

• max range usually at 45°

3.5 Projectiles Launched at an Angle

Height & Range

vup = –vdown

3.5 Projectiles Launched at an Angle

20 m/s

–20 m/s

12 m/s

12 m/s

12 m/s

12 m/s

10 m/s

12 m/s

–10 m/s

Velocity & Time

Is it safe to shoota bullet in the air?

0 m/s

3.5 Projectiles Launched at an Angle

Velocity & Time

tup = tdownvup = –vdown

ttotal = (2)tup

3.5 Projectiles Launched at an Angle

Height & Range

vx constant, but vy changes

At hmax, vy = 0 (only Vx horizontal)

Velocity & Time

tup = tdown

vup = –vdown

ttotal = (2)tup

more angle:

-more initial vy, more height

-less initial vx, less range

height

range

1. Without air resistance, the time for a vertically tossed ball to return to where it was thrown is …

A. 10 m/s for every second in the air.

B. the same as the time going upward.

C. less than the time going upward.

D. more than the time going upward.

Quick Quiz!

3.5 Projectiles Launched at an Angle

Solving projectile calculation problems in 3 easy steps:

1)Direction: get Vix & Viy (pick Horiz. or Vert.)

2)List Variables

d =

vi =

a =

vf =

t =

3. Pick equation, Plug numbers, and Solve.

3.5 Projectiles Launched at an Angle

Sample Calculation #1Bob Beamon’s record breaking long jump (8.9 m) at the 1968 Olympics resulted from an initial velocity of 9.4 m/s at an angle of 40o above horizontal.Solve for each of the following variables:

vix =

viy =

tup =

ttotal = (time of flight)

dx = (range)

dymax = (peak height)

vy = v sin(θ)

vx = v cos(θ)g = –10 m/s2

v = vi + at

d = vit + ½at2

ttotal = (2)tup

vix = (9.4 m/s) • cos(40o) =

viy = (9.4 m/s) • sin(40o) =

tup =

ttotal (up AND dn) = (2)(0.604 s) =

0 – 6.04 = –10

7.20 m/s

6.04 m/s

vfy = viy + at

0 = 6.04 + –10t0.604 s

1.21 s

Vi = 9.4 m/s

at 40o above horizontal 40o

9.4 m/s

dx =

dymax =

d = (7.20 m/s)(1.21 s)

d = vixt + ½at28.71 m0

d = (6.04 m/s)(0.604 s) + ½(–10 m/s2)(0.604 s)2 =

d = viyt + ½at21.82 m

Vo = 9.4 m/s

at 40o above x-axis 40o

9.4 m/s

vix =

viy =

7.20 m/s

6.04 m/s

tup =

ttotal =

0.604 s

1.21 s

7.20 m/s

6.04 m/s

0.604 s

1.21 s

8.71 m

1.82 m

3.5 Projectiles Launched at an Angle

A soccer ball is kicked horizontally off a 22.0 m high hill and lands a distance of 35.0 m from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

vix =

viy = 0 m/s

t =

dx = 35.0 m

dymax = 22.0 m

22.0 m

35.0 m

vix

Sample Calculation #2

3.5 Projectiles Launched at an Angle

viy = 0 m/s

dx = 35.0 m

dymax = 22.0 m

t =

vix =

22.0 m

35.0 m

vix

2(–22.0 m) = –10

d = viyt + ½at2

–22.0 = ½(–10)t2

0

√2.10 s

d = vixt + ½at2

35.0 = vix(2.10) 35.0 m = 2.10 s

016.7 m/s

Sample Calculation #2

3.5 Projectiles Launched at an Angle

θHorizontal Launchviy = 0 m/s

vix = v

Angled Launchviy = v sin(θ)

vix = v cos(θ)

vv

For ALL launches:

a = g = –10 m/s2 for vertical motion

a = 0 m/s2 for horizontal motion

t is found vertically with:

v = vi + gt or d = ½gt2